jmc

Drawing the long diagonals of a regular hexagon divides the hexagon into equilateral triangles with side length equal to half the length of each long diagonal. So, the area of the base equals 6 times the area of an equilateral triangle with side length 4. An equilateral triangle with side length 4 has area $4^2\sqrt{3}/4=4\sqrt{3}$, so the area of the base of the pyramid is $6(4\sqrt{3})=24\sqrt{3}$.

Let $O$ be the center of the hexagon, so $\overline{PO}$ is the altitude from the apex of the pyramid. Since triangle $PAD$ is an equilateral triangle, triangle $POA$ is a 30-60-90 triangle with hypotenuse 8. $\overline{PO}$ is opposite the $60^{\circ }$ angle in this triangle, so $PO=4\sqrt{3}$.

Finally, the volume of the pyramid is $$\frac{1}{3}\cdot [ABCDEF]\cdot PO=\frac{1}{3}\cdot 24\sqrt{3}\cdot 4\sqrt{3}=\boxed{96}.$$